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Math Help - Euler Identity Algebra Problem

  1. #1
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    Euler Identity Algebra Problem

    Using Euler's identity, find real positive constants c and \phi for all real t such that:

    3 \cos (2t) - 4 \sin (2t + \frac{\pi}{4}) = c \cos (2t + \phi)

    I tried substituting the Euler formula for all sine and cosine values to get:

    \frac{3}{2}(e^{j2t}+e^{-j2t}) - \frac{4}{2}(e^{j2t}e^{j\pi /4}-e^{-j2t}e^{-j\pi /4}) = \frac{c}{2}(e^{j2t}e^{j\phi}+e^{-j2t}e^{-j\phi})

    Which suggests:

    ce^{j\phi} = 3 - 4 e^{j\pi /4}

    Which I don't see a solution to. Any ideas?
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  2. #2
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    Re: Euler Identity Algebra Problem

    Quote Originally Posted by VinceW View Post
    Using Euler's identity, find real positive constants c and \phi for all real t such that:

    3 \cos (2t) - 4 \sin (2t + \frac{\pi}{4}) = c \cos (2t + \phi)

    I tried substituting the Euler formula for all sine and cosine values to get:

    \frac{3}{2}(e^{j2t}+e^{-j2t}) - \frac{4}{2}(e^{j2t}e^{j\pi /4}-e^{-j2t}e^{-j\pi /4}) = \frac{c}{2}(e^{j2t}e^{j\phi}+e^{-j2t}e^{-j\phi})

    Which suggests:

    ce^{j\phi} = 3 - 4 e^{j\pi /4}

    Which I don't see a solution to. Any ideas?
    \displaystyle \begin{align*} 3\cos{(2t)} - 4\sin{\left(2t + \frac{\pi}{4}\right)} &= 3\cos{(2t)} - 4\left[\sin{(2t)}\cos{\left(\frac{\pi}{4}\right)} + \cos{(2t)}\sin{\left(\frac{\pi}{4}\right)}\right] \\ &= 3\cos{(2t)} - 4\left[\frac{\sqrt{2}}{2}\sin{(2t)} + \frac{\sqrt{2}}{2}\cos{(2t)}\right] \\ &= 3\cos{(2t)} - 2\sqrt{2}\sin{(2t)} - 2\sqrt{2}\cos{(2t)} \\ &= \left(3 - 2\sqrt{2}\right)\cos{(2t)} - 2\sqrt{2}\sin{(2t)} \end{align*}

    And...

    \displaystyle \begin{align*} c\cos{\left(2t + \phi\right)} &= c\left[\cos{(2t)}\cos{(\phi)} - \sin{(2t)}\sin{(\phi)}\right] \\ &= c\cos{(\phi)}\cos{(2t)} - c\sin{(\phi)}\sin{(2t)}  \end{align*}

    So

    \displaystyle c\cos{(\phi)} = 3 - 2\sqrt{2} \\  \implies c^2\cos^2{(\phi)} = \left(3 - 2\sqrt{2}\right)^2 = 9 - 12\sqrt{2} + 8 = 17 - 12\sqrt{2}

    and \displaystyle c\sin{(\phi)} = 2\sqrt{2} \implies c^2\sin^2{(\phi)} = 8.

    Therefore

    \displaystyle \begin{align*} c^2\cos^2{(\phi)} + c^2\sin^2{(\phi)} &= 17-12\sqrt{2} + 2\sqrt{2} \\ c^2\left[\cos^2{(\phi)} + \sin^2{(\phi)} \right] &= 17 - 10\sqrt{2} \\ c^2(1) &= 17 - 10\sqrt{2} \\ c^2 &= 17 - 10\sqrt{2} \\ c &= \pm \sqrt{17 - 10\sqrt{2}} \end{align*}

    Can you go from here to get values for \displaystyle \phi?
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  3. #3
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    Re: Euler Identity Algebra Problem

    Thanks Prove It. You definitely solved it. You didn't use the Euler formula, but I guess that wasn't applicable?

    I believe you made a small algebra error in that this:

    c^2\cos^2{(\phi)} + c^2\sin^2{(\phi)} &= 17-12\sqrt{2} + 2\sqrt{2}

    should be:

    c^2\cos^2{(\phi)} + c^2\sin^2{(\phi)} &= 17-12\sqrt{2} + 8

    thanks
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  4. #4
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    Re: Euler Identity Algebra Problem

    Quote Originally Posted by VinceW View Post
    Thanks Prove It. You definitely solved it. You didn't use the Euler formula, but I guess that wasn't applicable?

    I believe you made a small algebra error in that this:

    c^2\cos^2{(\phi)} + c^2\sin^2{(\phi)} &= 17-12\sqrt{2} + 2\sqrt{2}

    should be:

    c^2\cos^2{(\phi)} + c^2\sin^2{(\phi)} &= 17-12\sqrt{2} + 8

    thanks
    Yes you are correct, that was a typo.

    Did you manage to find any values for \displaystyle \phi?
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  5. #5
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    Re: Euler Identity Algebra Problem

    Quote Originally Posted by Prove It View Post
    Did you manage to find any values for \displaystyle \phi?
    Yes, that was last step was obvious after all the previous steps:

    \phi = \arccos \frac{3 - 2\sqrt{2}}{\sqrt{25 - 12\sqrt{2}}} = \arcsin \frac{2\sqrt{2}}{\sqrt{25 - 12\sqrt{2}}} \approx 1.510
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